Final2

Name: Honor Code Signature:

Physics 101 Final Exam
8 Dec 2003 Prof L. Weinstein

There are 34 questions. Unless otherwise noted, please give a short explanation for all of your non-numerical answers. Show your work for all numerical answers.

$G = 6.67\cdot10^{-11}$ N $\cdot$ m /kg

$k = 9\cdot 10^9$ N $\cdot$ m /C

Earth's mass $M_e = 6\cdot 10^{24}$ kg

Earth's radius $R_e = 6.4\cdot 10^6$ m

Moon's mass $M_m = 7\cdot 10^{22}$ kg

Moon's radius $R_m = 1.7\cdot 10^6$ m

Earth-Moon distance $d_{E-m} = 3.8\cdot 10^8$ m

Earth-Sun distance $d_{E-S} = 1.5\cdot10^{11}$ m

I am not allowed to use any part of your SSN to publicly post grades or other private information without your permission.

Check here to let me use the last four digits of your student ID number to post your exam, homework, and lab grades on the web.

Alternatively, write a five digit number here for me to use to post your exam, homework, and lab grades on the web.

If two electrons initially at rest are placed close to each other they will exert a force on each other. Ignore all other forces. When the electrons are released and are free to move, they will move
1. further away from each other
2. not at all (they will not move)
3. closer to each other
4. Need more information
electrons are negatively charged and repel each other
In the previous problem, when the electrons are released and start moving, the force on each will
1. increase as they move
2. decrease as they move
3. remain constant as they move
4. Nonsense, they won't move
5. Need more information
The force between them is the COulomb force and is proportional to 1/d^2. As the distance increases, the force will decrease.
(no explanation needed) You electrically polarize a neutral metal sphere. This means that
1. the sphere is now electrically charged
2. some positive charges are on one side of the sphere and an equal number of negative charges are on the other side
3. it is now an insulator
4. it is magnetic
5. Need more information
(No explanation needed) I have two 100-g copper rods. I stretch one so that it is 10 cm long and about 1 cm in diameter. I stretch the other one so that it is 100 cm long and about 0.3 cm diameter. Which one has more resistance?
1. the 10 cm long rod
2. the 100 cm long rod
3. both are the same
4. Need more information
A 32-inch television set has about a 5000 V difference between the back of the set and the screen. The distance between the back of the TV set and the screen is 0.6 m. If 3 mC of electrons travel from the back of the set to the screen, how much kinetic energy do they gain?
The electrons are at a potential of 5000 V so they have potential energy PE = 3 mC * 5000 V = 3*10^(-3) C * 5000 V = 15 J This potential energy gets converted to kinetic energy so the KE = 15 J.
A 6-Volt battery is connected to a 3- $\Omega$ light bulb. What is the current flowing through the circuit?
I = V/R = 6 V/3 Ohms = 2 Amps
How much electrical power does the light bulb in the previous problem use?
P = IV = 2 Amps * 6 V = 12 Watts
If you double the voltage of the battery in the previous problem, what happens to the current in the circuit?
1. it quarters
2. it halves
3. it does not change
4. it doubles
5. it quadruples
6. Need more information
I = V/R. Doubling V will double I
A pair of light bulbs connected in parallel to a battery will draw
1. more current than a single bulb would draw
2. less current than a single bulb would draw
3. the same current that a single bulb would draw
4. Need more information
More current. Each light bulb in parallel draws as much current as the single bulb would. Therefore, two bulbs in parallel draw twice as much current. We say this in class when two bulbs were twice as bright as just one.
As you increase the number of identical light bulbs in this circuit from 1 to 2 to 3 as shown, the total power used by the circuit (if it is plugged into a 120-V AC wall outlet)
= 1.5in $\epsffile{seriesbulbs.eps}$
1. increases
2. decreases
3. remains the same
4. Need more information
These bulbs are in series since the same current must pass through all of them in series. Therefore, the total resistance of the circuit is the sum of the resistances and it increases as you add bulbs. Therefore the total current I = V/R decreases. The power P = IV therefore decreases. Also, we saw this in class that two bulbs in series give off less total light than one,
(no explanation needed) A moving electric charge will always interact with (ie: have a force exerted on it by)
1. a constant electric field
2. a constant magnetic field
3. both
4. neither
5. Need more information
An electric field will always exert a force on an electric charge (F = qE). A constant magnetic field will exert a force on an alectric charge if the charge is moving.
(No explanation needed) Like kinds of magnetic poles repel. Unlike kinds of magnetic poles
1. attract
2. repel
3. either attract or repel
4. Need more information
A beam of electrons is travelling toward the East. If I want to deflect the beam so it travels toward the north-east, then I should apply a force pointing
= 1.5in $\epsffile{beam1.eps}$
1. North
2. East
3. Up
4. West
5. other
6. Need more information
Vectors add. To make anything travelling East change direction and go NorthEast, you must apply a force in the North direction. There is no magnetic field in this problem.
A beam of electrons is travelling from East to West. If I want to apply a southward force, then I should apply a magnetic field pointing
= 1.5in $\epsffile{beam2.eps}$
1. North or South
2. East or West
3. Up or down (ie: into or out of the page)
4. nonsense, a magnetic field won't deflect the electron beam
5. Need more information
If a magnetic field exerts a southward force on a westward travelling electron beam, then the magnetic field must be perpendicular to both the direction of travel of the electrons and the force. Therefore, it must be perpendicular to the plane of the paper.
My particle detector needs about 12,000 V but only about 20 W to operate. I have a transformer with 50 turns in the primary coil. I connect the primary coil to a 120-V AC outlet and the secondary coil to my detector. How many turns does the transformer need to have in its secondary coil to supply 12,000 V to my detector?
The wattage is irrelevant. V(primary) / #Turns(primary) = V(secondary) so that
120/50 = 12,000/#turns so that
#turns = 50*12,000/120 = 5000
I make an electromagnet by winding an insulated wire around a hollow plastic tube. If I put 3 A of current through the wire, then the electromagnet can pick up two paperclips. Now I place an iron rod inside the plastic tube. How many paperclips can the electromagnet pick up now?
= 1.5in $\epsffile{electromagnet.eps}$
1. less than 2
2. 2
3. more than 2
4. Need more information
When you place some iron in an electromagnet, the magnetic field of the electromagnet will make the magnetic domains in the iron align themselves. This makes the magnetic field of the electromagnet much stronger. A shorter way to say it: Iron amplifies magnetic fields. Note that the iron itself is not a magnet. Placing the iron in the coil does not induce a voltage in the coil.
You do an experiment in outer space far away from the pesky effects of gravity and friction. A 0.2-kg bar magnet is launched horizontally from left to right at a speed of 4 m/s through a 0.5 m long stationary aluminum pipe with a mass of 1 kg. What is the speed of the magnet after it exits the pipe?
= 1.in $\epsffile{pipe.eps}$
1. less than 4 m/s
2. 4 m/s
3. more than 4 m/s
4. Need more information
We did this in class. When I dropped a magnet through the aluminum pipe, the magnet fell very slowly. There was a force between the magnet and the pipe that slowed down the magnet. In physics terms, as the magnet moves, the pipe sees a changing magnetic field. This changing magnetic field induces a voltage in the pipe that makes a current flow, that makes a magnetic field, that opposes the motion of the magnet (ie: repels the magnet).
In the previous problem, in what direction is the pipe moving after the magnet exits the pipe?
1. to the left (opposite the direction of the magnet)
2. to the right (in the same direction as the magnet)
3. nonsense, the pipe is not moving
4. Need more information
The pipe applies a force on the magnet to the left (that slows down the magnet). Therefore, the magnet applies a force on the pipe to the right (Newton's 3rd Law). Therefore the pipe moves to the right.

Another way to look at it is that the momentum of the pipe plus magnet is conserved. Therefore, if the magnet slows down, the pipe must take some of that speed to make up for it (and conserve momentum).
(no explanation needed) I use a bar magnet to pick up a small iron rod. This temporarily magnetizes the rod. Which diagram indicates the magnetic orientation of the rod?
= 1.in $\epsffile{magnet.eps}$
1. N on top, S on bottom
2. S on top, N on bottom
3. N on top, N on bottom
4. S on top, S on bottom
5. it is not magnetized at all
6. Need more information
A large wheel is free to rotate around its hub. Three ropes are attached to three points on the wheel. You can pull on any one rope with a certain force for a very short time (same force and time on each rope). Which rope should you pull on in order to exert the maximum torque on the wheel? (You cannot change the direction of the rope.)
= 1.in $\epsffile{wheel.eps}$
1. rope A
2. rope B
3. rope C
4. rope A or C
5. Need more information
Torque = force times lever arm. The lever arm is the perpendicular distance from the axis (or hub) to the point the force is applied. C has a lever arm of zero, B has a lever arm of about 1/2 and A has a lever arm of 1. Another way to look at it is that if you pull on rope C, you won;t make the wheel turn, therefore you are not exerting any torque.
The wheel shown below has a mass of 30 kg. I add a 5-kg block to the wheel. Where should I add the block in order to maximize the rotational inertia of the wheel plus block?
= 1.in $\epsffile{wheel2.eps}$
1. point A
2. point B
3. point C
4. points A or C
5. Need more information
rotational inertia = mr^2. An object is harder to turn if the mass is farther from the axis.
A ball slides down a curved frictionless slope as shown in the figure. At which point does the ball have the greatest acceleration?
= 1.in $\epsffile{slope.eps}$
1. point A
2. point B
3. point C
4. all the same
5. Need more information
At point A, the ball is falling almost straight down. Therefore it is accelerating with almost 10 m/s^2. The fact that its velocity at point A is very small is irrelevant. At point B, the slope is no longer vertical so the acceleration is smaller. At point C, the track is horizontal so there is no more acceleration. In the absence of friction, the ball will move with constant speed after point C.
A ball slides rolls down a curved frictionless slope as shown in the previous problem. At which point does the ball have the greatest speed?
1. point A
2. point B
3. point C
4. all the same
5. Need more information
At point C, all the potential energy has been turned into kinetic energy and so the ball is moving at its fastest there. Another way to look at it is that at point C, it has fallen the farthest so it will be going the fastest.
Dave's pickup truck has rear-wheel drive. In order to improve traction in the snow, Dave put three fifty-pound sandbags in the back of the pickup truck over the rear wheels. The sandbags are not tied down. If the pickup truck stops VERY suddenly, where will the sandbags end up?
1. They will move backwards (relative to the pickup) and fall off the rear of the pickup truck
2. They will stay where they are in the pickup truck
3. They will move forwards (relative to the pickup) and slam into the cabin (front) of the pickup truck
4. Need more information
Newton's 1st Law (an object in motion tends to stay in motion). If you had read the comics that morning, you would have seen the answer to this problem here
An 5-kg object is sliding in a straight line along the surface of an icy pond at a constant 13 m/s. What is the net force on the object?

Zero.
A proton (mass = 1 amu) travelling at m/s hits a neutron of the same mass. They stick together to form a deuterium nucleus.
1. What is their total momentum before the collision? (Hint: use `amu' for the mass units instead of kg.)
  
  p = m1*v1 + m2*v2 = (1 amu)*(2*10^7 m/s) + (1 amu)*(0) = 2*10^7 amu m/s
2. What is the speed of the deuterium nucleus (ie: the proton and neutron stuck together) after the collision?
  
  p(after) = p(before)
  m(after)*v(after) = p(before) = 2*10^7 amu m/s
  (2 amu)*v(after) = 2*10^7 amu m/s
  v(after) = 1*10^7 m/s
Please note that part b of the problem asked for the speed, not the momentum.
Suppose that there is another planet named `Dirt' that is the same size as the Earth but with only half the mass. The gravitational acceleration on the surface of Dirt will be
1. 1/4 that of Earth
2. 1/2 that of Earth
3. the same as Earth
4. twice that of Earth
5. four times that of Earth
6. Need more information
The gravitational acceleration on the surface of a planet is g = GM/d^2. Since Dirt has the same size as Earth, d is the same. Since M is half as large, g will be half as large.
A satellite in orbit 200 km above the Earth needs a speed of 8 km/s to stay in orbit. A satellite in orbit around Dirt will need a speed of
1. less than 8 km/s
2. 8 km/s
3. more than 8 km/s
4. Need more information
There will be less gravitational acceleration, therefore the satellite will not fall as far in 1 sec (ie: it will fall less than 5 m). Therefore, it does not have to travel as fast to stay in orbit.
A satellite is in orbit around the Earth at a height of 200 km (6600 km from the center of the Earth). It has a tangential speed of 8 km/s. How much time does it take to complete one orbit around the Earth?
This is simply time = distance/speed = circumference/speed = 2 pi r / speed = 2 * pi * 6600 km / 8 km/s = 5200 sec = 1.4 hours.

This is the amount of time it takes a satellite (or the International Space Station) to orbit the Earth.
A gallon of gasoline contains about J of chemical energy. If we could convert that chemical energy to kinetic energy with 100% efficiency, what speed would your 2000 kg car be going (ignore friction and air resistance)? Give your answer in m/s and then connvert it to mph using 1 m/s $\approx 2$ mph. Is this a lot?
KE = 1/2 mv^2 and m = 2000 kg (a LARGE SUV) so we have
10^8 J = 1/2 * (2000 kg) * v^2
v^2 = 10^8 J / 1000 kg = 10^5 m^2/s^2
v = sqrt(10^5) m/s = 300 m/s
This is approximately 600 mph.
Wow, that's fast.
You raise a textbook of mass to certain height, giving it a certain amount of potential energy. If you now raise a second textbook of twice the mass to the same height, its potential energy will be
1. 1/4 as much
2. 1/2 as much
3. the same
4. twice as much
5. four times as much
6. Need more information
PE = mgh. Double m will double PE.
Two charged objects that are free to move are separated by a distance of 10 m. Object has a mass kg and a charge mC. Object has a mass kg and a charge mC. What is the electrical force acting on object ?
Coulomb's law: F = k * q1 * q2 / d^2 = 9*10^9 Nm^2/C^2 * (0.004 C) * (0.008 C) / (10 m)^2 = 2.88*10^3 N
I took one point off for not knowing that mC = milliCoulomb = 10^(-3) C and I took one point off for having the wrong units for force.
In the previous problem, what is the electrical force acting on object ?

F = 2.88*10^3 N. Newton's 3rd Law. No matter how wrong your answer was to the previous question, if you had the same answer for this one, I gave you full credit.
In the previous problem, what is the acceleration of object ?

a = F/m = 2.88*10^3 N / 3 kg = 960 m/s^2

2003-12-10